Biharmonic functions on Schrödinger networks

Abstract

In an infinite graph X,  the equation $\Delta u\left( x\right) =\varphi \left( x\right) u\left( x\right) ,\varphi \left( x\right) \ge 0,$ is termed the Schrödinger equation. The solutions of this equation are named $\varphi -$harmonic functions. This article studies varied aspects associated with these solutions: Harnack principle, minimum principle, domination principle, Dirichlet solution etc. The main thrust is the development of an abstract theory of discrete $\varphi -$biharmonic functions on X and use it for the $\varphi -$biharmonic classification of infinite graphs.

1 Introduction

A function theory on finite graphs based on potential-theoretic concepts is useful in the study of electrical networks satisfying Ohm’s law and Kirchhoff’s current–voltage laws. An extended study on infinite graphs is warranted as it has its applications too as suggested by the search for approximate solutions of differential equations related to wavesas well as in the study of recurrent and transient random walks.

A function theory on an infinite network, made up of an infinite graph and a set of transition functions can be developed as a discrete model for the study of subharmonic functions on the Euclidean spaces. Thus, the discrete versions of the Laplace operator Green functions, logarithmic functions, potential functions etc. are considered on an infinite network. On the Euclidean spaces, we define $\Delta _{q}$ as a generalised Schrödinger operator when $q\left( x\right)$ is a locally (Lebesgue) integrable function and $\Delta _{q}u\left( x\right) =\Delta u\left( x\right) -q\left( x\right) u\left( x\right)$ in the sense of distributions for any locally (Lebesgue) integrable function $u\left( x\right)$. The case usually studied is when $q\left( x\right)$ is constant and $\Delta$ is the Laplacian operator in the classical sense. A discrete version of $\Delta _{q}$ where $q\left( x\right)$ is a real-valued function on an infinite graph is discussed in this paper. For related studies we cite (Abodayeh [1]; Ficher 2021; Manivannan (to appear); Nathiya [10]) on finite and infinite graphs.

In the classical theory of time-independent Schrödinger operators in the Euclidean spaces, the perturbed Laplace equation $\Delta u\left( x\right) =\mu u\left( x\right)$ where $\mu >0$ is a constant, plays an important rôle. The solutions and supersolutions of this equation have many properties akin to harmonic and superharmonic functions defined on the domains in the complex plane and in Riemannian manifolds. To understand and develop the inter-relation among these properties like Harnack principle, minimum principle, domination principle, Dirichlet solution to some boundary-value problems etc., it is found useful to consider the solutions to the equation $\Delta u\left( x\right) =\varphi \left( x\right) u\left( x\right)$ in a discrete set-up, where the real-valued function $\varphi \left( x\right) \ge 0.$

Accordingly, in this article we take up the study of the discrete equation $\Delta u\left( x\right) =\varphi \left( x\right) u\left( x\right) ,$ where $\varphi \left( x\right) \ge 0$ is a real-valued function on an infinite graph X,  with a countable number of vertices and a countable number of edges. Some of the properties of the solutions and the supersolutions of this equation, named here as $\varphi$-harmonic and $\varphi$-superharmonic functions on X,  are already known and a few others are proved here. Based on them, the thrust of this article is to develop a theory of $\varphi$-biharmonic and $\varphi$-bisuperhamonic functions on X.

When $\varphi =0$ and X is the Euclidean space, biharmonic functions appear in the context of elasticity of thin plates and associated problems. Here we present an abstract theory of $\varphi$-biharmonic functions in the discrete case and initiate a classification theory of infinite graphs depending on the existence of $\varphi$-biharmonic functions with varied characteristics. It is found convenient occasionally to use the terms from the study of random walks while presenting the results.

2 Preliminaries

In an infinite graph X with a countably infinite number of vertices and a countable infinite number of edges, two vertices x and y are said to be neighbors, $x\sim y$, if there is an edge joining x and y; two vertices a, b are connected if there exists a path $\{a=x_{0},x_{1},...,x_{n}=b\},$ $x_{i}\sim x_{i+1},$ connecting a and b;  the graph X is connected if any two vertices in X are connected by a path; it is locally finite if any vertex has only a finite number of neighbors; there is no loop in X if for any vertex x,  $x\sim x$ is not valid.

A transition set $\{t\left( x,y\right) \}$ in X is a set of numbers $t(x,y)\ge 0$ such that $t(x,y)>0$ if and only if $x\sim y;$ the numbers $t\left( x,y\right) ,t\left( y,x\right)$ need not be the same. An infinite network $\left\{ X,t\left( x,y\right) \right\}$ consists of an infinite graph X that is connected, locally finite and without loops, provided with a transition set $\{t\left( x,y\right) \}.$ Write $p\left( x,y\right) =\frac{t\left( x,y\right) }{t\left( x\right) }$ where $t\left( x\right) =\sum _{y\sim x}t\left( x,y\right) .$ We refer to the infinite network $\left\{ X,p\left( x,y\right) \right\}$ as the random walk associated to $\left\{ X,t\left( x,y\right) \right\} .$

In a random walk X, given a real-valued function $u\left( x\right) ,$ define the averaging operator A by $Au\left( x\right) =\sum _{y\sim x}p\left( x,y\right) u\left( y\right) .$ For a fixed real-valued function $\varphi \left( x\right) \ge 0,\varphi \not \equiv 0,$ define the Schrödinger operator $A^{^{\prime }}$ by $A^{^{\prime }}u\left( x\right) =\sum _{y\sim x}p^{^{\prime }}\left( x,y\right) u\left( y\right)$ where $p^{^{\prime }}\left( x,y\right) =\frac{p\left( x,y\right) }{1+\varphi \left( x\right) }.$

2.1 The relevance of the operators A and $A^{^{\prime }}$ is as follows:

• (1) In $\left\{ X,t\left( x,y\right) \right\} ,$ the Laplace operator $\Delta$ is given by

  $$\begin{aligned} \Delta u(x)&=\sum _{y}t\left( x,y\right) [u(y)-u(x)]=-t(x)u(x)+\sum _{y}t(x,y)u(y)\\&=t(x)[-u(x)+\sum _{y}p(x,y)u(y)]=t(x)[(A-I)u(x)]. \end{aligned}$$

  Consequently, $\Delta u\left( x\right) \le 0$ if and only if $Au\left( x\right) \le u\left( x\right) .$ If we say that $u\left( x\right)$ is superharmonic on X if $Au\left( x\right) \le u\left( x\right) ,$ then it is the same as saying $\Delta u(x)\le 0.$ This shows that from a potential theory perspective (that is, statements concerning superharmonic, harmonic functions, potentials etc. and their properties), the network $\left\{ X,t\left( x,y\right) \right\}$ behaves as its associated random walk $\left\{ X,p\left( x,y\right) \right\} .$

• (2) In $\left\{ X,t\left( x,y\right) \right\} ,$ given a real-valued function $q\left( x\right) \ge 0,q\not \equiv 0.$ The operator $\Delta _{q}u(x)=\Delta u(x)-q(x)u(x)$ is known as the Schrödinger operator. Thus

  $$\begin{aligned} \Delta _{q}u(x)&=-[t(x)+q(x)]u(x)+\sum _{y}t(x,y)u(y)\\&=t(x)\{\sum _{y}p(x,y)u(y)\}-[1+\frac{q\left( x\right) }{t\left( x\right) }]u(x)\}\\&=t(x)[1+\varphi (x)]\left\{ \sum _{y}p^{^{\prime }}(x,y)u(y)-u(x)\right\} \end{aligned}$$

  where $\varphi (x)=\frac{q\left( x\right) }{t\left( x\right) }.$Hence $\Delta _{q}u(x)=t(x)\left[ 1+\varphi \left( x\right) \right] [(A^{^{\prime }}-I)u(x)].$ Consequently $\Delta _{q}u\left( x\right) \le 0$ if and only if $A^{^{\prime }}u(x)\le u(x).$ We say that u(x) is $\varphi$-superharmonic if $A^{^{\prime }}u\left( x\right) \le u\left( x\right) .$

Definition 1

A real-valued function $u\left( x\right)$ is superharmonic at x if and only if $Au\left( x\right) \le u\left( x\right) ,$ subharmonic at x if and only if $Au\left( x\right) \ge u\left( x\right)$ and harmonic at x if and only if $Au\left( x\right) =u\left( x\right) .$ A superharmonic function $s\left( x\right) \ge 0$ is a potential if we have $u(x)\le 0$ for any subharmonic function $u\left( x\right) \le s\left( x\right) .$ Similar definitions for $\varphi$-superharmonic functions with respect to the operator $A^{^{\prime }}.$

Any non-negative $\varphi$-subharmonic function is subharmonic, any non-negative $\varphi$-superharmonic function is superharmonic; as a result, any potential is a $\varphi$-potential.

Definition 2

A vertex x is said to be an interior vertex of a set E if x and all its neighbors are in E.

Denoting all the interior points of E by $\overset{0}{E}$, define the boundary $\partial E=E\setminus \overset{\circ }{E}.$

Definition 3

A real-valued function $u\left( x\right)$ on E is said to be $\varphi$-superharmonic on E,  if $A^{^{\prime }}u(x)\le u(x)$ for any $x\in \overset{\circ }{E}.$

2.2 Some properties of $\varphi$-superharmonic functions

• (1) If u and v are $\varphi$-superharmonic on E,  then $\inf (u,v)$ is $\varphi$-superharmonic on E.

• (2) If $\left\{ u_{n}\right\}$ is a sequence of $\varphi$-superharmonic functions on E and if $\lim _{n}u_{n}(x)=u(x)$ exists and is real-valued, then $u\left( x\right)$ is $\varphi$-superharmonic on E.

• (3) (Poisson $\varphi$-modification): If u(x) is $\varphi$-superharmonic on E and if $z\in \overset{\circ }{E,}$ then define the function v(x) on E such that $v(x)=u(x)$ if $x\ne z$ and $v\left( x\right) =\sum _{y}p^{^{\prime }}\left( z,y\right) u\left( y\right) .$ Then v(x) is $\varphi$-superharmonic on $E,v\left( x\right)$ is $\varphi$-harmonic at $x=z,$ and $v\left( x\right) \le u\left( x\right)$ on E. The function $v\left( x\right)$ is referred to as the Poisson $\varphi$-modification of $u\left( x\right)$ at $x=z.$

• (4) (The greatest $\varphi$-harmonic minorant): Let $u\left( x\right) \ge v\left( x\right)$ on E where u(x) is $\varphi$-superharmonic and $v\left( x\right)$ is $\varphi$-subharmonic. Let ${\mathcal {F}}$ be the family of all $\varphi$-superharmonic functions $s\left( x\right)$ on E where $s\left( x\right) \ge v\left( x\right) .$ Then $\inf _{s\in {\mathcal {F}} }s\left( x\right) =h\left( x\right)$ is a $\varphi$-harmonic function on E;  moreover, if $h^{^{\prime }}\left( x\right)$ is any $\varphi$-harmonic function on E such that $h^{^{\prime }}\left( x\right) \ge v\left( x\right) ,$ then $h^{^{\prime }}\left( x\right) \ge h\left( x\right)$ also.

Proof

Since the family ${\mathcal {F}}$ is lower directed and since E is a countable set, then there exists a decreasing sequence $s_{n}\left( x\right) \in {\mathcal {F}}$ such that

$$\begin{aligned} \inf _{s\in {\mathcal {F}} }s\left( x\right) =\lim _{n}s_{n}\left( x\right) =h\left( x\right) \end{aligned}$$

which is $\varphi$-superharmonic on E. Now for any $z\in \overset{\circ }{E,}$ let $h_{1}\left( x\right)$ be the Poisson $\varphi$-modification of $h\left( x\right)$ at $x=z.$ Then $h_{1}\in {\mathcal {F}}$ so that $h_{1}\ge h.$ But by the property of the Poisson $\varphi$-modification $h_{1}\le h.$ Hence $h_{1}=h$ which implies that h(x) is $\varphi$-harmonic at $x=z.$ Since z is an arbitrary interior vertex of E,  we conclude that h(x) is $\varphi$-harmonic on E. Moreover, if $h^{^{\prime }}$ is any $\varphi$-harmonic function such that $h^{^{\prime } }\ge v,$ then $h^{^{\prime }}\in {\mathcal {F}} \$and hence $h^{^{\prime }}\ge h.$ $\square$

• 5. Since $\varphi \not \equiv 0,$ then the constant 1 is $\varphi$-superharmonic, but not $\varphi$-harmonic on$\ X.$ Let h be the greatest $\varphi$-harmonic minorant of 1. If 1 is not a $\varphi$-potential, then $\ 1-h$ is a $\varphi$-potential. Thus in any case there are $\varphi$-potentials on X.

• (5) (Minimum Principle): If $s\left( x\right)$ is a $\varphi$-superharmonic function on a finite subset E and if $s\left( x\right) \ge 0$ on $\partial E,$ then $s\ge 0$ on E.

Proof

Suppose s(x) takes negative values on E,  let $\inf _{x\in E}s\left( x\right) =-\alpha <0.$ Then for some $z\in \overset{\circ }{E},s\left( z\right) =-\alpha .$ Since $s\left( z\right) \ge \sum _{y}p^{^{\prime } }(z,y)s(y),$ then $\sum _{y}p^{^{\prime }}(z,y)\left[ s\left( y\right) +\alpha \right] \le s\left( z\right) +\alpha \sum _{y}p^{^{\prime } }(z,y)=\alpha \left[ \sum _{y}p^{^{\prime }}(z,y)-1\right] \le 0.$ Since each term on the left side is non-negative, then $s\left( y\right) =-\alpha ,$ for each $y\sim z.$ This is a contradiction if one such neighbor of z is on $\partial E.$ Otherwise repeat the argument with each neighbor of z until we arrive at a contradiction. Hence $\inf _{x\in E}s\left( x\right) \ge 0.$ Hence $s\ge 0$ on E. $\square$

• 7. (Dirichlet Problem): Let f(x) be a real-valued function defined on the boundary $\partial E$ of a finite set E. Then there exists a unique $\varphi$-harmonic function $h\left( x\right)$ on E such that $h(x)=f(x)$ on $\partial E.$

Proof

Since E is a finite set, choose some $M>0$ such that $-M<f(x)<M$ on $\partial E.$ Then the function $s\left( x\right)$ on E such that $s\left( x\right) =f\left( x\right)$ on $\partial E$ and $s\left( x\right) =M$ on $\overset{\circ }{E}$ is $\varphi$-superharmonic. Let $\aleph$ be the family of all $\varphi$-superharmonic functions $u\left( x\right)$ on E such that $u\left( x\right) =f\left( x\right)$ on $\partial E.$ Note that each $u\left( x\right) \ge -M$ on E by the Minimum Principle. Then $h(x)=\inf _{u\in \aleph }u(x)$ is a bounded $\varphi$-harmonic function on E such that $h(x)=f(x)$ on $\partial E.$ The uniqueness of the solution h(x) is a consequence of the Minimum Principle. $\square$

• 8. ($\varphi$- Green Function): For any vertex z,  there exists a unique $\varphi$-potential $G_{z}^{^{\prime }}$ such that $(A^{^{\prime }}-I)G_{z}^{^{\prime }}(x)=-\delta _{z}\left( x\right)$ for all $x\in X.$

Proof

Let $\aleph$ be the family of all $\varphi$-superharmonic functions $u\left( x\right)$ such that $u(x)\ge \delta _{z}(x).$ Note that the constant function $1\in \aleph$. Let $s\left( x\right) =\inf _{u\in \aleph }u(x),$ then $s\left( x\right)$ is $\varphi$-superharmonic.

By using the Poisson $\varphi$-modification locally, $A^{^{\prime }}s\left( x\right) \le s\left( x\right)$ on X and $A^{^{\prime }}s\left( x\right) =s\left( x\right)$ if $x\ne z.$ Consequently $A^{^{\prime } }s\left( x\right) -s\left( x\right) =\alpha \delta _{z}\left( x\right)$ for some $\alpha <0.$ Take $G_{z}^{^{\prime }}(x)=-\frac{1}{\alpha }s\left( x\right) .$ Then $(A^{^{\prime }}-I)G_{z}^{^{\prime }}(x)=-\delta _{z}\left( x\right) .$

To see the uniqueness of $G_{z}^{^{\prime }}(x):$ Suppose $g(x)>0$ is another $\varphi$-potential such that $(A^{^{\prime }}-I)g(x)=-\delta _{z}\left( x\right) .$ Then $G_{z}^{^{\prime }}(x)-g(x)=h\left( x\right)$ which is a $\varphi$-harmonic function. Since $h(x)\le G_{z}^{^{\prime }}(x)$ and $-h(x)\le g(x),$ then $h=0.$ $\square$

• 9. (Domination Principle): Let $u\left( x\right)$ be a $\varphi$-potential which is $\varphi$-harmonic at each vertex in E. If $s\left( x\right) >0$ is a $\varphi$-superharmonic function on X such that $s\left( x\right) \ge u\left( x\right)$ on $X{\setminus } E,$ then $s\ge u$ on X.

Proof

Let $v\left( x\right) =\inf \left\{ s\left( x\right) ,u\left( x\right) \right\} .$ Then $v\left( x\right)$ is a $\varphi$-potential on X. Therefore $f\left( x\right) =u\left( x\right) -v\left( x\right)$ is a $\varphi$-subharmonic function on X with values 0 on $X{\setminus } E.$ Since $f\left( x\right) \le u\left( x\right)$ on X,  then $f\le 0,$ hence $u\left( x\right) \le v\left( x\right) .$ Thus $u\left( x\right) =v\left( x\right)$ showing that $u\left( x\right) \le s\left( x\right)$ on X. $\square$

Corollary 1

Let z be a vertex in X,  $G_{z}^{^{\prime }}\left( x\right)$ the $\varphi$-Green function with pole at z. Then,

• (a) $G_{z}^{^{\prime }}\left( x\right) \le G_{z}^{^{\prime }}\left( z\right) .$

• (b) If the Green function $G_{z}^{^{{}}}\left( x\right)$(that is the potential $G_{z}^{^{{}}}\left( x\right)$ such that $\left( I-A\right) G_{z}^{{}}\left( x\right) =\delta _{z}\left( x\right)$) exists on X,  then $G_{z}^{{}}\left( x\right)$ is a $\varphi$-superharmonic function on X,  so that $G_{z}^{^{\prime }}\left( x\right) \le \frac{G_{z}^{^{\prime } }\left( z\right) }{G_{z}^{{}}\left( z\right) }G_{z}^{{}}\left( x\right) .$

• (c) $G_{z}^{{}}\left( x\right) =G_{z}^{^{\prime }}\left( x\right) +v_{z}\left( x\right)$ where $v_{z}\left( x\right)$ is a $\varphi$-potential. Consequently every potential in X is the sum of two $\varphi$-potentials.

Proof

$\left( A-I\right) G_{z}^{{}}\left( x\right) =-\delta _{z}\left( x\right)$ and $\left( A^{^{\prime }}-I\right) G_{z}^{^{\prime }}\left( x\right) =-\delta _{z}\left( x\right)$

$\left( A^{^{\prime }}-I\right) G_{z}^{{}}\left( x\right) \le \left( A-I\right) G_{z}^{{}}\left( x\right) =\left( A^{^{\prime }}-I\right) G_{z}^{^{\prime }}\left( x\right) .$ Hence

$\left( A^{^{\prime }}-I\right) \left[ G_{z}^{{}}\left( x\right) -G_{z}^{^{\prime }}\left( x\right) \right] \le 0.$

That is, $G_{z}^{{}}\left( x\right) -G_{z}^{^{\prime }}\left( x\right) =v_{z}\left( x\right)$ which is $\varphi$-superharmonic. Since

$-v_{z}\left( x\right) \le G_{z}^{^{\prime }}\left( x\right) ,$ then $-v_{z}\left( x\right) \le 0.$ Thus $0\le v_{z}(x)\le G_{z}^{^{{}}}\left( x\right)$ which shows that $v_{z}\left( x\right)$ is a $\varphi$-potential. Now let s(x) be a potential in X. Then $s\left( x\right) =\sum _{y}\alpha \left( y\right) G_{y}^{{}}\left( x\right)$ where the constant $\alpha \left( y\right) =(I-A)s(y)\ge 0.$

$s(x)=\sum _{y}\alpha (y)G_{y}^{^{\prime }}\left( x\right) +\sum _{y} \alpha (y)v_{y}(x),$ the sum of two $\varphi$-potentials. $\square$

3 $\varphi$-Biharmonic functions and $\varphi$-potentials

Many of the important properties of polyharmonic functions in ${\mathbb {R}} ^{n}$ are given in Nicolesco (1930).

Among other papers related to biharmonic functions we mention a few:

• (1) Sario et al. (1977) develop a biharmonic classification theory of Riemannian manifolds analogous to the harmonic classification theory of Riemann surfaces.

• (2) Yamasaki (1980) gives a discrete version of biharmonic functions on infinite graphs with symmetric transition functions.

• (3) Smyrnelis [14] and [15] presents a development of positive bisuperharmonic functions in the Brelot-Bauer axiomatic set-up on a locally compact space.

• (4) The probabilistic interpretation of biharmonic functions is given in Bouleau [5].

• (5) In this context, it is interesting to see the paper Boboc, Bucur [4] in the continuous case identifying positive bisuperharmonic functions with the excessive functions of a triangular resolvent.

Definition 4

Let u and $v\$be real-valued functions on X such that $(I-A^{^{\prime } })u(x)=v(x).$ Then we say that

• (1) u is $\varphi$-bisuperharmonic ($\varphi$-biharmonic) if v is $\varphi$-superharmonic ($\varphi$-harmonic).

• (2) u is $\varphi$-bipotential if u and v are $\varphi$-potentials.

• (3) u is $\varphi$-biharmonic potential if u is $\varphi$-potential and v is $\varphi$-harmonic.

Parallel definitions for bisuperharmonic, biharmonic, bipotentials and biharmonic potentials for the operator $(I-A).$

Proposition 1

If $\{u_{n}\}$ is a sequence of $\varphi$-bisuperharmonic (biharmonic) functions and if $\lim _{n}u_{n}\left( x\right) =u\left( x\right)$ exists and is real-valued for each x in X,  then $u\left( x\right)$ is $\varphi$-bisuperharmonic ($\varphi$-biharmonic).

Proof

Let $(I-A^{^{\prime }})u_{n}=v_{n}.$ Then $v_{n}$ is $\varphi$-superharmonic ($\varphi$-harmonic). Since $\lim _{n\rightarrow \infty }u_{n}=u$ exists, then $\lim _{n}v_{n}=\lim _{n}\left( I-A^{^{\prime }}\right) u_{n}=\left( I-A^{^{\prime }}\right) u.$ Since $v_{n}$ is $\varphi$-superharmonic ($\varphi$-harmonic) and since $\lim _{n}v_{n}=v$ exists, then v is $\varphi$-superharmonic ($\varphi$-harmonic). Thus $\left( I-A^{^{\prime }}\right) u=v,$ and u is $\varphi$-bisuperharmonic ($\varphi$-biharmonic). $\square$

The constant 1 is $\varphi$-superharmonic but not $\varphi$-harmonic, hence there are $\varphi$-potentials on X. But the existence of $\varphi$-bipotentials on X is not assured. A necessary and sufficient condition for the existence of $\varphi$-bipotentials is: there exists a positive $\varphi$-superharmonic function $s\left( x\right)$ such that $\sum _{y}s\left( y\right) G_{y}^{^{\prime }}\left( x\right)$ is real-valued for at least one vertex x. Some of the sufficient conditions for this existence are:

Proposition 2

If there is a bipotential on X,  then there exist $\varphi$-bipotentials on X.

Proof

Let $w\left( x\right)$ be a bipotential on X. That is, $\left( I-A\right) w\left( x\right) =s\left( x\right)$ where $w\left( x\right)$ and $s\left( x\right)$ are potentials. Since $w\left( x\right)$ is a potential,

$$\begin{aligned} w\left( x\right)&=\sum _{y}\left[ \left( I-A\right) w\left( y\right) \right] G_{y}\left( x\right) \\&=\sum _{y}s\left( y\right) G_{y}\left( x\right) \\&\ge \sum _{y}s\left( y\right) G_{y}^{^{\prime }}\left( x\right) =u\left( x\right) . \end{aligned}$$

Hence $u\left( x\right)$ is a $\varphi$-potential such that $\left( I-A^{^{\prime }}\right) u\left( x\right) =s\left( x\right)$ where $s\left( x\right)$ is a potential, hence a $\varphi$-potential on X. In other words, $u\left( x\right)$ is a $\varphi$-bipotential on X. $\square$

Notation 1

Let $R_{h}^{e}(x)$ stand for the infimum of all positive $\varphi$-bisuperharmonic functions $s\left( x\right)$ on X such that $s\left( e\right) \ge h\left( e\right) .$

Proposition 3

If there is a $\varphi$-biharmonic potential, then there are $\varphi$-bipotentials on X.

Proof

Let $u\left( x\right)$ be a $\varphi$-biharmonic potential. Then $\left( I-A^{^{\prime }}\right) u\left( x\right) =h\left( x\right)$ is a $\varphi$-harmonic function. Since $u\left( x\right)$ is a $\varphi$-potential, then $h(x)\ge 0.$ For a vertex e,  let $v(x)=R_{h}^{e}(x)$ Then $v\left( x\right)$ is a $\varphi$-potential and $v\left( x\right) \le h\left( x\right) .$ Now $u(x)=\sum _{y}h(y)G_{y}^{^{\prime }}(x)\ge \sum _{y}v(y)G_{y}^{^{\prime }}(x)=w(x)$ which is a $\varphi$-potential. Hence $(I-A^{^{\prime }})w(x)=v(x).$ Therefore $w\left( x\right)$ is a $\varphi$-bipotential on X. $\square$

Remark 1

Replace the set $\{e\}$ by a finite set E in the proof above. Then: If there is a $\varphi$-biharmonic potential on X,  then there exists a $\varphi$-bipotential which is $\varphi$-biharmonic outside E.

Notation 2

When $f\left( x\right) ,g\left( x\right)$ are two real-valued functions on X,  write $f\succcurlyeq g$ to denote $f\left( x\right) \ge g\left( x\right)$ and $(I-A^{^{\prime }})f\left( x\right) \ge (I-A^{^{\prime } })g\left( x\right) .$

The following proposition shows that if there is a $\varphi$-bisuperharmonic function $u\succeq v$ (that is not $\varphi$-harmonic), then there exists $\varphi$-bipotential on X.

Proposition 4

If $u\succcurlyeq 0$ is a $\varphi$-bisuperharmonic function, then u is the unique sum of a $\varphi$-bipotential, a $\varphi$-biharmonic function and a non-negative $\varphi$-harmonic function.

Proof

Since $u\ge 0$ and $(I-A^{^{\prime }})u=v\ge 0,$ then u is a $\varphi$-superharmonic function and hence the unique sum of a $\varphi$-potential s, and a non-negative $\varphi$-harmonic function h. Then $s\left( x\right) =\sum _{y}v(y)G_{y}^{^{\prime }}(x).$ Now $v\ge 0$ is also a $\varphi$-superharmonic function, hence $v=\left( \varphi -\text {potential }w\right) +$(a $\varphi$-harmonic function $H\ge 0$). Consequently $s\left( x\right) =\sum _{y}w(y)G_{y}^{^{\prime }}(x)+\sum _{y}H\left( y\right) G_{y}^{^{\prime } }\left( x\right) =\mu \left( x\right) +\sigma \left( x\right)$ where $\mu \left( x\right)$ is a $\varphi$-bipotential and $\sigma \left( x\right)$ is a $\varphi$-biharmonic potential. Finally, $u(x)=s(x)+h(x)=\mu (x)+\sigma (x)+h(x).$

The uniqueness of this decomposition follows from the uniqueness of decomposition of a non-negative $\varphi$-superharmonic function as the sum of a $\varphi$-potential and a non-negative $\varphi$-harmonic function. $\square$

Remark 2

(Representation of a biharmonic function $b\succeq 0$). Let $b\left( x\right) \succeq 0$ be $\varphi$-biharmonic. That is, $(I-A^{^{\prime }})b\left( x\right) =h\left( x\right) \ge 0.$ Since $(I-A^{^{\prime }})b\left( x\right) \ge 0,$ then $b\left( x\right)$ is a positive $\varphi$-superharmonic function, hence the sum of a $\varphi$-potential and a non-negative $\varphi$-harmonic function. Thus, $b\left( x\right) =\sum _{y}h\left( y\right) G_{y}^{^{\prime }}\left( x\right) +v\left( x\right)$ where $v\left( x\right) \ge 0$ is $\varphi$-harmonic. Now a representation of the non-negative $\varphi$-harmonic function $v\left( x\right)$ can be obtained by using the minimal $\varphi$-harmonic functions, leading to a Choquet integral representation. In this context, the more inclusive paper by Picardello and Woess [12] describing boundary representation of polyharmonic functions is of interest.

Proposition 5

Let s be a $\varphi$-bisuperharmonic function and u be a $\varphi$-potential such that $0\preceq s\preceq u.$ Then s is a $\varphi$-bipotential.

Proof

Since $0\le s\le u$ and $0\le (I-A^{^{\prime }})s\le (I-A^{^{\prime }})u,$ both s and $\left( I-A^{^{\prime }}\right) s$ are $\varphi$-potentials, hence s is a $\varphi$-bipotential on X. $\square$

The assumption on the density function $\varphi \left( x\right)$ is $\varphi \ge 0,\not \equiv 0.$ Under these general conditions, there may not be any positive $\varphi$-bipotentials on X. ( In the classical Schrödinger operator case, $\varphi \left( x\right)$ is a positive constant). However,

Proposition 6

If $\varphi \left( x\right) \ge \varepsilon >0,$ then there exist $\varphi$-potentials on X.

Proof

The constant function 1 is $\varphi$-superharmonic. Note $(I-A^{^{\prime } })[1(x)]=1-\sum _{y}p^{^{\prime }}(x,y)[1(y)]=1-\frac{1}{1+\varphi \left( x\right) }\ge 1-\frac{1}{1+\varepsilon }=\frac{\varepsilon }{1+\varepsilon }.$

$1=\sum _{y}\left( I-A^{^{\prime }}\right) [1(y)G_{y}^{^{\prime }}(x)+$(a non–negative $\varphi$-harmonic function $H\left( x\right) ],$ then $1\ge \sum _{y}\frac{\varepsilon }{1+\varepsilon }G_{y}^{^{\prime }}\left( x\right) .$ Hence $s(x)=\sum _{y}G_{y}^{^{\prime }}(x)$ is a $\varphi$-potential such that $(I-A^{^{\prime }})s=1.$That is, $s\succcurlyeq 0$ is a $\varphi$-bisuperharmonic function, hence $\varphi$-bipotentials exist on X. $\square$

Remark 3

In the above proof, if the constant 1 is not a $\varphi$-potential, then $0<H(x)<1$ is a bounded $\varphi$-harmonic function. Since $\sum _{y} G_{y}^{^{\prime }}(x)$ is real valued, then $\sum _{y}H\left( y\right) G_{y}^{^{\prime }}(x)=v\left( x\right)$ is a $\varphi$-potential. Since $\left( I-A^{^{\prime }}\right) v\left( x\right) =H\left( x\right) ,$ then $v\left( x\right)$ is a bounded $\varphi$-biharmonic potential on X.

Proposition 7

Suppose $\varphi \left( x\right) \ge \varepsilon >0.$ If there exists a bounded $\varphi$-biharmonic function on X,  then there exist bounded $\varphi$-biharmonic potentials (hence bounded $\varphi$-bipotentials also) on X.

Proof

Let $\left( I-A^{^{\prime }}\right) u\left( x\right) =v\left( x\right) ,$ where $v\left( x\right)$ is a $\varphi$-harmonic function and $\left| u\left( x\right) \right| \le M.$ Since $\left| \left( I-A^{^{\prime }}\right) u\left( x\right) \right| =\left| u\left( x\right) -\sum _{y}p^{^{\prime }}(x,y)u\left( y\right) \right| \le M+M\sum _{y}p^{^{\prime }}(z,y)\le 2M,$ then the $\varphi$-subharmonic function $\left| v\left( x\right) \right| \le 2M;$ since the constant 2M is $\varphi$-superharmonic, then there exists a $\varphi$-harmonic function $h\left( x\right)$ such that $\left| v\left( x\right) \right| \le h\left( x\right) \le 2M.$ Since $\sum _{y}G_{y}^{^{\prime }}(x)$ is real-valued, bounded, then $s\left( x\right) =\sum _{y}h(y)G_{y}^{^{\prime } }(x)$ is a $\varphi$-potential such that $\left( I-A^{^{\prime }}\right) s\left( x\right) =h\left( x\right) ;$ that is $s\left( x\right)$ is a bounded $\varphi$-biharmonic potential on X. $\square$

Note: If there are no positive potentials on X,  then the constant 1 is a $\varphi$-potential. For, if 1 is not $\varphi$-potential, then there exists a bounded $\varphi$-harmonic function $H\left( x\right) ,0<H(x)<1.$ This function $H\left( x\right)$ is then subharmonic on X,  which means that $H\left( x\right)$ is a constant since there are no positive potentials on X. This assertion on $H\left( x\right)$ is not valid, leading to the conclusion that 1 is a $\varphi$-potential.

Theorem 1

($\varphi$-biharmonic Green potentials) Suppose there exist $\varphi$-bipotentials on X. Let z be an arbitrary vertex in X. Then there exists the unique function $Q_{z}^{^{\prime }}\left( x\right)$ such that $\left( I-A^{^{\prime }}\right) Q_{z}^{^{\prime }}\left( x\right) =G_{z}^{^{\prime }}\left( x\right) ;$ and $Q_{z}^{^{\prime }}\left( x\right)$ is $\varphi$-biharmonic in $X{\setminus }\left\{ z\right\} .$ The $\varphi$-potential $Q_{z}^{^{\prime }}\left( x\right)$ is referred to as the $\varphi$-biharmonic Green potential with pole at z.

Proof

Take a $\varphi$-bipotential $u\left( x\right)$ on $X,(I-A^{^{\prime } })u\left( x\right) =v(x).$ Then

$$\begin{aligned} u(x)=\sum _{y}v(y)G_{y}^{^{\prime }}(x). \end{aligned}$$

Since $v\left( x\right)$ is a $\varphi$-potential, by the Domination Principle, $v\left( x\right) \ge \rho G_{z}^{^{\prime }}\left( x\right)$ for a constant $\rho >0.$ Then, $u(x)=\sum _{y}v(y)G_{y}^{^{\prime }}(x)\ge \rho \sum _{y}G_{z}^{^{\prime }}(y)G_{y}^{^{\prime }}(x)=\rho Q_{z}^{^{\prime } }\left( x\right) ,$ where $Q_{z}^{^{\prime }}\left( x\right) =\sum _{y} G_{z}^{^{\prime }}(y)G_{y}^{^{\prime }}(x)$ is a $\varphi$-potential such that $(I-A^{^{\prime }})Q_{z}^{^{\prime }}(x)=G_{z}^{^{\prime }}(x).$ Hence $Q_{z}^{^{\prime }}\left( x\right)$ is a $\varphi$-bipotential that is $\varphi$-biharmonic in $X\setminus \left\{ z\right\} .$ The uniqueness proof is standard. $\square$

Proposition 8

If there exist bipotentials on X and if $(I-A)Q_{z}(x)=G_{z}(x),$ then $Q_{z}(x)-Q_{z}^{^{\prime }}(x)=s(x)$ is a $\varphi$-potential.

Proof

Since $(I-A^{^{\prime }})Q_{z}(x)\ge (I-A)Q_{z}(x)=G_{z}(x)\ge G_{z} ^{^{\prime }}(x)=(I-A^{^{\prime }})Q_{z}^{^{\prime }}(x),$ then $(I-A^{^{\prime } })[Q_{z}(x)-Q_{z}^{^{\prime }}(x)]\ge 0.$ Hence $Q_{z}(x)-Q_{z}^{^{\prime } }(x)=s(x),$ which is a $\varphi$-superharmonic function. Now $-s(x)\le Q_{z}^{^{\prime }}\left( x\right) .$ Therefore $-s(x)\le 0.$ Suppose $v\left( x\right)$ is a $\varphi$-subharmonic function such that $0\le v(x)\le s(x)$. Then $v(x)\le Q_{z}(x),$ where $v\left( x\right)$ is subharmonic (since a non-negative $\varphi$-subharmonic function is subharmonic) and $Q_{z}\left( x\right)$ is a potential and so $v\le 0.$ Consequently, $s\left( x\right)$ is a $\varphi$-potential. $\square$

4 Biharmonic extension

Bajunaid et al. [3] consider the representation of discrete biharmonic functions defined outside a finite set in an infinite tree T without terminal vertices and without any positive potentials defined on it. A useful result is that given any real-valued function $f\left( x\right)$ on T,  one could always construct a function $g\left( x\right)$ such that $\left( I-A\right) g\left( x\right) =f\left( x\right)$ on T.

(An infinite tree is an infinite graph without any closed path; that is for any vertex y there is no path $\left\{ y,y_{1},...,y_{n},y\right\} ,n\ge 2$). However in the present case of $\varphi$-biharmonic functions on a Schrödinger network X there are always positive $\varphi$-potentials on X. Bajunaid [2] shows that in a hyperbolic Riemannian manifold R for any given biharmonic function $b\left( x\right)$ outside a compact set there exists a biharmonic function $B\left( x\right)$ on R such that $B\left( x\right) -b\left( x\right)$ is bounded outside a compact set. In Bajunaid et al. [3], there is a note on the probabilistic interpretation of discrete biharmonic functions.

Theorem 2

($\varphi$-biharmonic extension) Let b(x) be a $\varphi$-biharmonic function defined outside a finite set. Then outside a finite set $b(x)=v(x)+s(x),$ where $v\left( x\right)$ is a uniquely determined $\varphi$-biharmonic function on X and $s\left( x\right)$ is the difference of two $\varphi$-potentials on X that are $\varphi$-biharmonic outside a finite set and hence bounded on X.

Proof

For a finite set E in X,  let $V(E)=E\cup \{z\in X,z$ has a neighbor in $E\}.$ Let the $\varphi$-biharmonic function $b\left( x\right)$ be defined outside the finite set F. Take a finite set E such that $E^{\circ }$ includes F. Let h(x) be $\varphi$-harmonic on E which is the $\varphi$-Dirichlet solution with boundary value $b\left( x\right)$ on $\partial E.$ Define f(x) on X such that $f(x)=h(x)$ on E,  and $f(x)=b(x)$ on $X\backslash E.$ Note that $X\backslash V(E)\subset (X\backslash E)^{\circ }.$ For if $z\in X\backslash V(E)$ and if $y\sim z,$ then $y\notin V(E).$ Thus z and all its neighbors are in $X\backslash E,$ that is $z\in (X\backslash E)^{\circ }.$ Take now on X,  the function

$$\begin{aligned} v(x)=f(x)-\sum _{y\in V\left( E\right) \setminus E^{\circ }}[(I-A^{^{\prime } })f(y)]G_{y}^{^{\prime }}(x). \end{aligned}$$

• (1) If $x\in E^{\circ },$ then $(I-A^{^{\prime }})v(x)=0.$

• (2) If $z\in V\left( E\right) \setminus E^{\circ },$ then

  $$\begin{aligned} (I-A^{^{\prime }})v\left( z\right)&=(I-A^{^{\prime }})f\left( z\right) -\sum _{y\in V\left( E\right) \setminus E^{\circ }}[(I-A^{^{\prime }})f(y)] \left[ (I-A^{^{\prime }})G_{y}^{^{\prime }}(z)\right] \\&=(I-A^{^{\prime }})f\left( z\right) -\sum _{y\in V\left( E\right) \setminus E^{\circ }}[(I-A^{^{\prime }})f(y)]\left[ \delta _{y}\left( z\right) \right] \\&=(I-A^{^{\prime }})f(z)-\left[ (I-A^{^{\prime }})f\left( z\right) \right] \\&=0. \end{aligned}$$

• (3) If $x\notin V(E),$ then $v(x)=b(x)-s(x)$ where

  $$\begin{aligned} s\left( x\right) =\sum _{y\in V\left( E\right) \setminus E^{\circ } }[(I-A^{^{\prime }})f(y)]G_{y}^{^{\prime }}(x), \end{aligned}$$

  which is the difference of two $\varphi$-potentials on X and $\varphi$-harmonic at every vertex outside $V\left( E\right) {\setminus } E^{\circ }.$ Moreover, since $G_{y}^{^{\prime }}(x)\le G_{y}^{^{\prime }}(y)$ for any x in X then $\left| s\left( x\right) \right| \le \sum _{y\in V\left( E\right) {\setminus } E^{\circ }}\left| (I-A^{^{\prime }})f(y)\right| G_{y}^{^{\prime }}(x),$ which is a bounded $\phi -$potential on X. Consequently, v(x) is a $\varphi$-biharmonic function on X such that $b(x)=v(x)+s(x)$ outside a finite set where the bounded function $\left| s\left( x\right) \right|$ is bounded by a $\varphi$-potential.

Uniqueness of the $\varphi$-biharmonic function v(x) :  Suppose $b_{1} (x)=v_{1}(x)+s_{1}(x)$ is another such decomposition. Then $v_{1}(x)-v\left( x\right) =s(x)-s_{1}(x)$ outside a finite set. Consequently, the $\varphi$-harmonic function $(I-A^{^{\prime }})(v_{1}-v)$ on X is 0 outside a finite set, and so $(I-A^{^{\prime }})(v_{1}-v)=0$ on X. Hence $v_{1}-v=h$ is a $\varphi$-harmonic function on X. Consequently, the $\varphi$-subharmonic function |h| is bounded by a $\varphi$-potential on X,  leading to the conclusion $h=0.$ $\square$

Corollary 2

If h(x) is a $\varphi$-harmonic function outside a finite set, then there exists a unique $\varphi$-harmonic function $v\left( x\right)$ on X such that $h(x)=v(x)+s(x)$ outside a finite set, where $s\left( x\right)$ is the difference of two $\varphi$-potentials on X that are $\varphi$-harmonic outside a finite set. Hence $s\left( x\right)$ is bounded on X.

Proof

Consider h(x) as a $\varphi$-biharmonic function outside a finite set. Then by the above theorem, $h(x)=v(x)+s(x)$ outside a finite set. Then $(I-A^{^{\prime }})v(x)=0$ outside a finite set. Consequently, the $\varphi$-biharmonic function h(x) is actually $\varphi$-harmonic on X. $\square$